Question

data Set a = Set [a]
i1 = Set [1, 2, 3]
i2 = Set [3, 2, 1]

member xs x  = elem x xs
subset xs ys = and (map (member ys) xs)

instance (Eq a) => Eq (Set a) where
  (Set xs) == (Set ys) = (subset xs ys) && (subset ys xs)

class Eq a => Comparable a where
  cmp :: a -> a -> String
  cmp x y
    | x == y    = "eq"
    | otherwise = "neq"

instance Comparable a => Comparable (Set a) where
  cmp (Set xs) (Set ys)
    | (Set xs) == (Set ys) = "same"
    | otherwise            = "different"

执行时：cmp i1 i2我收到以下错误：

No instance for (Comparable Integer) arising from a use of ‘cmp’  
 In the expression: cmp i1 i2  
 In an equation for ‘it’: it = cmp i1 i2

我想知道这意味着什么？感谢。

Answer 1

执行时：

cmp i1 i2

Haskell将i1和i2的类型推断为Set [Integer]。由于cmp的类型为：

cmp :: (Comparable a) => a -> a -> String

Integer必须有Comparable类型的实例;它没有。因此，要解决此问题，您应为Comparable提供Integer的实例。

另请注意，您可能需要使用Ord来定义不同类型的值之间的比较。

什么＆＃34;没有（可比较的整数）＆＃34;意思？

1 个答案: